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2 years ago

Add me on discord if u wanna be study buddies or just normal friends lol

Chemistry

2 answers:

Katyanochek1 [597]2 years ago

7 0

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beks73 [17]2 years ago

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Sure what the discord

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20 poll

aleksley [76]

Answer: n=15.56moles

Explanation:

PV = nRT

where

P is pressure in atmospheres

V is volume in Liters

n is the number of moles of the gas

R is the ideal gas constant = given as (0.0821L -atm/k-mol

PV = nRT

n= PV/RT

n= (1.5 X 230)/ (0.0821 X 270)

n= 15.56 moles

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2 years ago

The water cycle imagine you are a rain drop

Soloha48 [4]

Evaporation,condensation,precipitation,sublimation,transpirtation,runoff and infiltration

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2 years ago

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valina [46]

2)c

3)b

4)c

5)a

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2 years ago

Read 2 more answers

1- Which process are fast in carbon cycle?

g100num [7]

Answer:

During photosynthesis, plants absorb carbon dioxide and sunlight to create fuel—glucose and other sugars—for building plant structures. This process forms the foundation of the fast (biological) carbon cycle.

The Slow Carbon Cycle. ... Atmospheric carbon combines with water to form a weak acid—carbonic acid—that falls to the surface in rain. The acid dissolves rocks—a process called chemical weathering—and releases calcium, magnesium, potassium, or sodium ions.

6 0

2 years ago

The solubility of silver(I)phosphate at a given temperature is 1.02 g/L. Calculate the Ksp at this temperature. After you get yo

Snezhnost [94]

<u>Answer:</u> The solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

<u>Explanation:</u>

We are given:

Solubility of silver (I) phosphate = 1.02 g/L

To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:

Molar mass of silver (I) phosphate = 418.6 g/mol

$\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L$

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The chemical equation for the ionization of silver (I) phosphate follows:

$Ag_3PO_4(aq.)\rightleftharpoons 3Ag^{+}(aq.)+PO_4^{3-}(aq.)$

3s s

The expression of $K_{sp}$ for above equation follows:

$K_{sp}=(3s)^3\times s$

We are given:

$s=2.44\times 10^{-3}M$

Putting values in above expression, we get:

$K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}$

Hence, the solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

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2 years ago